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Unformatted text preview: Kim, Jin Homework 5 Due: Oct 2 2007, 3:00 am Inst: Diane Radin 1 This printout should have 17 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Determine y when y = x x . 1. y = xy (2 ln x + 1) 2. y = xy (2 ln x + 1) 3. y = xy (2 ln x 1) 4. y = y x 2 (ln x 1) 5. y = y (ln x + 1) 6. y = y (ln x + 1) correct 7. y = y x 2 (ln x 1) 8. y = y (2 ln x 1) Explanation: After taking natural logs we see that ln y = x ln x. Thus by implicit differentiation, 1 y dy dx = ln x + 1 . Consequently, y = y (ln x + 1) . keywords: 002 (part 1 of 1) 10 points If g = f , determine d dx ln(2 x 2 + f (6 x 2 )) . 1. x n 2 + 6 g (6 x 2 ) 2 x 2 + f (6 x 2 ) o 2. 4 x + 12 xg (6 x 2 ) (2 x 2 + f (6 x 2 )) 2 3. 2 x n 2 + 6 g (6 x 2 ) 2 x 2 + f (6 x 2 ) o correct 4. 12 xg (6 x 2 ) 2 x 2 + f (6 x 2 ) 5. 2 + 6 g (6 x 2 ) 2 x 2 + f (6 x 2 ) Explanation: By the Chain Rule d dx ln(2 x 2 + f (6 x 2 )) = 4 x + 12 xf (6 x 2 ) 2 x 2 + f (6 x 2 ) . Thus d dx ln(2 x 2 + f (6 x 2 )) = 2 x n 2 + 6 g (6 x 2 ) 2 x 2 + f (6 x 2 ) o . keywords: 003 (part 1 of 1) 10 points Find the absolute minimum of f on the interval [5 , 8] when f ( x ) = 4 n x 5 ln x 2 5 o 20 . 1. abs. min = 10 . 21 2. abs. min = 6 Kim, Jin Homework 5 Due: Oct 2 2007, 3:00 am Inst: Diane Radin 2 3. abs. min = 7 4. abs. min = 8 correct 5. abs. min = 8 . 35 Explanation: As f is differentiable everywhere on (2 , ), its absolute minimum on the interval [5 , 8] will occur at an endpoint of [5 , 8] or at a local minimum of f in (5 , 8). Now f ( x ) = 4 n 1 5 x 2 o and f 00 ( x ) = 20 ( x 2) 2 . So x = 7 will be a critical point at which f ( x ) will have a local minimum. But f (5) = 12 4 n 8 ln 3 5 o = 10 . 21 , f (7) = 8 , f (8) = 4 n 3 5 ln 6 5 o = 8 . 35 . Thus, on [5 , 8], abs. min = 8 . keywords: 004 (part 1 of 1) 10 points Determine the indefinite integral I = Z 6 x ( x 3) 2 dx. 1. I = 18 ( x 3) 2 + C 2. I = ln( x 3) 6 18 x 3 + C correct 3. I = ln( x 3) 6 + 18 ( x 3) 2 + C 4. I = 3 ln( x 3) 2 + C 5. I = 6 x 3 + C Explanation: Set u = x 3 ; then du = dx , so I = 6 Z x ( x 3) 2 dx = 6 Z ( u + 3) u 2 du = 6 Z du u + 18 Z u 2 du. But 6 Z du u = 6 ln  u  + C = ln u 6 + C, while 18 Z u 2 du = 18 u 1 + C. Consequently, I = ln( x 3) 6 18 x 3 + C ....
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This note was uploaded on 04/12/2010 for the course PHY 58195 taught by Professor Turner during the Spring '09 term at University of Texas at Austin.
 Spring '09
 Turner

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